Find the value of $p$ for which the function $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x \neq 0 \\ p, & x = 0 \end{cases}$ is continuous at $x = 0$.

8

The correct answer is Option (2) → 8 ##

$\lim\limits_{x \to 0} f(x) = f(0)$

$\lim\limits_{x \to 0} \frac{4 \times 2 \sin^2 2x}{4x^2} = p$

$8 \lim\limits_{x \to 0} \left( \frac{\sin 2x}{2x} \right)^2 = p$

$⇒p = 8$